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For an annual function, an organizer needs to select some participants
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24 Jun 2025, 02:11
24 Jun 2025, 02:11
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For an annual function, an organizer needs to select some participants, which should be evenly divisible in the groups of 9,10 , and 12 . What is the minimum number of participants the organizer can select?
Show: ::
180
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Re: For an annual function, an organizer needs to select some participants
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24 Jun 2025, 06:00
24 Jun 2025, 06:00
1
The concept being tested is for LCM. LCM of 9, 10 and 12 is 180.
Re: For an annual function, an organizer needs to select some participants
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25 Jun 2025, 04:15
25 Jun 2025, 04:15
Expert Reply
To find the minimum number of participants that can be evenly divided into groups of 9,10 , and 12, we need to determine the Least Common Multiple (LCM) of these numbers. The LCM is the smallest number that is a multiple of all three.
Step-by-Step Solution:
1. Prime Factorization:
$\(9=3^2\)$
$\(10=2 \times 5\)$
$\(12=2^2 \times 3\)$
2. Identify the Highest Exponents for Each Prime Factor:
2: The highest power is $2^2$ (from 12).
3: The highest power is $3^2$ (from 9).
5: The highest power is 5 (from 10).
3. Calculate the LCM:
$$
\(\mathrm{LCM}=2^2 \times 3^2 \times 5=4 \times 9 \times 5=180\)
$$
Verification:
- $\(180 \div 9=20\)$ (divisible)
- $\(180 \div 10=18\)$ (divisible)
- $\(180 \div 12=15\)$ (divisible)
Final Answer:
180
Step-by-Step Solution:
1. Prime Factorization:
$\(9=3^2\)$
$\(10=2 \times 5\)$
$\(12=2^2 \times 3\)$
2. Identify the Highest Exponents for Each Prime Factor:
2: The highest power is $2^2$ (from 12).
3: The highest power is $3^2$ (from 9).
5: The highest power is 5 (from 10).
3. Calculate the LCM:
$$
\(\mathrm{LCM}=2^2 \times 3^2 \times 5=4 \times 9 \times 5=180\)
$$
Verification:
- $\(180 \div 9=20\)$ (divisible)
- $\(180 \div 10=18\)$ (divisible)
- $\(180 \div 12=15\)$ (divisible)
Final Answer:
180
